How do you solve #secx+5=2secx+3# for #0<=x<=2pi#?

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Answer 1 · 2018-04-11T08:26:13

#color(blue)(pi/3, (5pi)/3)#

Identity:

#color(red)bb(secx=1/cosx)#

#secx+5=2secx+3#

Collect like terms:

#2secx-secx=5-3#

#secx=2#

Using identity:

#1/cosx=2#

#cosx=1/2#

#x=arccos(cosx)=arccos(1/2)=>x=pi/3#

This is in the I quadrant. The cosine is positive so, we also have an angle in the IV quadrant:

#2pi-pi/3=(5pi)/3#

So our solutions for:

#0 <= x <= 2pi#

#color(blue)(pi/3, (5pi)/3)#